Problem:
You have a symmetric parabola shape like this, and you are trying to compute the area, what do you do?
Solution:
To generalize the problem let me denote $ b $ as the base and $ h $ as the height instead of those specific numbers
The first thing to do is to find the equation. It passes through $ (-\frac{b}{2}, 0) $ and $ (\frac{b}{2}, 0) $ suggest that it has the form $ y = s(x - \frac{b}{2})(x + \frac{b}{2}) = s(x^2 - \frac{b^2}{4}) $
It passes through $ (0, h) $ suggest that $ -s\frac{b^2}{4} = h $ or $ s = -\frac{4h}{b^2} $.
So the full equation of the parabola is $ -\frac{4h}{b^2}x^2 + h $.
The area can be found as
$ \begin{eqnarray*} A &=& \int\limits_{-\frac{b}{2}}^{\frac{b}{2}}(-\frac{4h}{b^2}x^2 + h)dx \\ &=& -\frac{4h}{3b^2}x^3 + hx |_{-\frac{b}{2}}^{\frac{b}{2}} \\ &=& (-\frac{4h}{3b^2}(\frac{b}{2})^3 + h(\frac{b}{2})) - (-\frac{4h}{3b^2}(-\frac{b}{2})^3 + h(-\frac{b}{2})) \\ &=& (-\frac{hb}{6} + \frac{hb}{2}) - (\frac{hb}{6} - \frac{hb}{2}) \\ &=& \frac{2hb}{3} \end{eqnarray*} $
This last formula is really simple and worth remembering. A triangle is $ \frac{hb}{2} $, and a parabola is $ \frac{2hb}{3} $.
You have a symmetric parabola shape like this, and you are trying to compute the area, what do you do?
Solution:
To generalize the problem let me denote $ b $ as the base and $ h $ as the height instead of those specific numbers
The first thing to do is to find the equation. It passes through $ (-\frac{b}{2}, 0) $ and $ (\frac{b}{2}, 0) $ suggest that it has the form $ y = s(x - \frac{b}{2})(x + \frac{b}{2}) = s(x^2 - \frac{b^2}{4}) $
It passes through $ (0, h) $ suggest that $ -s\frac{b^2}{4} = h $ or $ s = -\frac{4h}{b^2} $.
So the full equation of the parabola is $ -\frac{4h}{b^2}x^2 + h $.
The area can be found as
$ \begin{eqnarray*} A &=& \int\limits_{-\frac{b}{2}}^{\frac{b}{2}}(-\frac{4h}{b^2}x^2 + h)dx \\ &=& -\frac{4h}{3b^2}x^3 + hx |_{-\frac{b}{2}}^{\frac{b}{2}} \\ &=& (-\frac{4h}{3b^2}(\frac{b}{2})^3 + h(\frac{b}{2})) - (-\frac{4h}{3b^2}(-\frac{b}{2})^3 + h(-\frac{b}{2})) \\ &=& (-\frac{hb}{6} + \frac{hb}{2}) - (\frac{hb}{6} - \frac{hb}{2}) \\ &=& \frac{2hb}{3} \end{eqnarray*} $
This last formula is really simple and worth remembering. A triangle is $ \frac{hb}{2} $, and a parabola is $ \frac{2hb}{3} $.
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